Solution Found!
Answer: CALC A cutting tool under microprocessor control
Chapter 7, Problem 83P(choose chapter or problem)
A cutting tool under microprocessor control has several forces acting on it. One force is
\(\vec{F}=a x y^{2} \hat{\jmath}\) a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{~N} / \mathrm{m}^{3}\). Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{~m}\) and \(y=3.00 \mathrm{~m}\).
(a) Calculate the work done on the tool by \(\vec{F}\) if this displacement is along the straight line that connects these two points.
(b) Calculate the work done on the tool by \(\vec{F}\)if the tool is first moved out along the \(x \text {-axis }\) to the point \(x=3.00 \mathrm{~m}, y=0\) and then moved parallel to the \(y \text {-axis }\) to the point \(x=3.00 \mathrm{~m}, y=3.00 \mathrm{~m}\) (c) Compare the work done by \(\vec{F}\) along these two paths. Is \(\vec{F}\) conservative or nonconservative? Explain.
Equation Transcription:
Text Transcription:
\vec{F}=a x y^{2} \hat{\j}
\alpha=2.50 \{~N} / \{m}^{3}
x=3.00 m
y=3.00 m
Vec F
y=x
Vec F
x-axis
x=3.00m, y=0
y-axis
x=3.00 m, y=3.00 m
Vec F
Vec F
Questions & Answers
QUESTION:
A cutting tool under microprocessor control has several forces acting on it. One force is
\(\vec{F}=a x y^{2} \hat{\jmath}\) a force in the negative y-direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{~N} / \mathrm{m}^{3}\). Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{~m}\) and \(y=3.00 \mathrm{~m}\).
(a) Calculate the work done on the tool by \(\vec{F}\) if this displacement is along the straight line that connects these two points.
(b) Calculate the work done on the tool by \(\vec{F}\)if the tool is first moved out along the \(x \text {-axis }\) to the point \(x=3.00 \mathrm{~m}, y=0\) and then moved parallel to the \(y \text {-axis }\) to the point \(x=3.00 \mathrm{~m}, y=3.00 \mathrm{~m}\) (c) Compare the work done by \(\vec{F}\) along these two paths. Is \(\vec{F}\) conservative or nonconservative? Explain.
Equation Transcription:
Text Transcription:
\vec{F}=a x y^{2} \hat{\j}
\alpha=2.50 \{~N} / \{m}^{3}
x=3.00 m
y=3.00 m
Vec F
y=x
Vec F
x-axis
x=3.00m, y=0
y-axis
x=3.00 m, y=3.00 m
Vec F
Vec F
ANSWER:
Solution 83P
Step 1 of 7:
Here in the given problem, we need to calculate the work done by given variable force along two different paths and comparing work done along two different paths, we need to comment on the conservative property of force.
Given data,
Force,
Force constant=2.5 N/m
To find ,
Work done , W=?
The work done by variable force along the path from point to
is given by,
W=
Where F is applied force and dr is displacement vector.